Question

1. Solve the quadratic equation by factorisation method: y2 - 7y + 12 = 0.​

Answer 1

AnswEr-:

• $\underline{\boxed {\mathrm {\blue{ The\:value\:of\:y\:is\:3\:or\:4.} }}}$

Explanation-:

• Solve this quadratic equation using Factorisation method -: y² - 7y + 12 .

$\dag{\mathrm { Solution \:of\: Question-:}}$

• $\longrightarrow {\mathrm { y^{2} - 7y + 12=0 }}$

$\sf{\dag{ By\:Using \:Sum-\:Product \:pattern-:}}$

$\underline{\sf{ ( Since\: -: (-7) = -3 + (-4) \: and \:(-3) \times (-4) =12 ) }}$

• $\longrightarrow {\mathrm { y^{2} - \purple{7y} + 12=0 }}$

• $\longrightarrow {\mathrm { y^{2} \purple {-4y - 3y } +   12=0 }}$

$\sf{ Now\:Finding \:Common-\:Factors \:from\:each\:term-:}$

• $\longrightarrow {\mathrm { y^{2} - 4y - 3y +   12=0 }}$

$\sf{ By\:Taking\:y\:as \:common\:in\:First \:term-:}$

• $\longrightarrow {\mathrm { \purple{y^{2} - 4y} - 3y +   12=0 }}$

• $\longrightarrow {\mathrm { \purple{y(y - 4)} - 3y +   12=0 }}$

$\sf{ By\:Taking\:-3\:as \:common\:in\:Second \:term-:}$

• $\longrightarrow {\mathrm { y(y - 4) \purple {- 3y +   12}=0 }}$

• $\longrightarrow {\mathrm { y(y - 4) \purple {- 3( y -  4)}=0 }}$

$\sf{ Now,\:Rewrite \:in\:the\:factored\:term\:\:-:}$

• $\longrightarrow {\mathrm {\purple {y(y - 4) - 3( y -  4)}=0 }}$

• $\longrightarrow {\mathrm {\purple {(y - 3) ( y -  4) } =0 }}$

As , We know that ,

• $\longrightarrow {\mathrm {(y - 4) =0 }}$

• $\longrightarrow {\mathrm {y =4 }}$

And ,

• $\longrightarrow {\mathrm {(y - 3) =0 }}$

• $\longrightarrow {\mathrm {y =3 }}$

Hence ,

• $\underline{\boxed {\mathrm {\blue{ The\:value\:of\:y\:is\:3\:or\:4.} }}}$

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